我想在JavaScript中使用XMLHttpRequest发送一些数据。

假设我在HTML中有以下表单:

<form name="inputform" action="somewhere" method="post">
  <input type="hidden" value="person" name="user">
  <input type="hidden" value="password" name="pwd">
  <input type="hidden" value="place" name="organization">
  <input type="hidden" value="key" name="requiredkey">
</form>

如何在JavaScript中使用XMLHttpRequest编写等效内容?


当前回答

var util = {
    getAttribute: function (dom, attr) {
        if (dom.getAttribute !== undefined) {
            return dom.getAttribute(attr);
        } else if (dom[attr] !== undefined) {
            return dom[attr];
        } else {
            return null;
        }
    },
    addEvent: function (obj, evtName, func) {
        //Primero revisar attributos si existe o no.
        if (obj.addEventListener) {
            obj.addEventListener(evtName, func, false);

        } else if (obj.attachEvent) {
            obj.attachEvent(evtName, func);
        } else {
            if (this.getAttribute("on" + evtName) !== undefined) {
                obj["on" + evtName] = func;
            } else {
                obj[evtName] = func;
            }

        }

    },
    removeEvent: function (obj, evtName, func) {
        if (obj.removeEventListener) {
            obj.removeEventListener(evtName, func, false);
        } else if (obj.detachEvent) {
            obj.detachEvent(evtName, func);
        } else {
            if (this.getAttribute("on" + evtName) !== undefined) {
                obj["on" + evtName] = null;
            } else {
                obj[evtName] = null;
            }
        }

    },
    getAjaxObject: function () {
        var xhttp = null;
        //XDomainRequest
        if ("XMLHttpRequest" in window) {
            xhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        return xhttp;
    }

};

//START CODE HERE.

var xhr = util.getAjaxObject();

var isUpload = (xhr && ('upload' in xhr) && ('onprogress' in xhr.upload));

if (isUpload) {
    util.addEvent(xhr, "progress", xhrEvt.onProgress());
    util.addEvent(xhr, "loadstart", xhrEvt.onLoadStart);
    util.addEvent(xhr, "abort", xhrEvt.onAbort);
}

util.addEvent(xhr, "readystatechange", xhrEvt.ajaxOnReadyState);

var xhrEvt = {
    onProgress: function (e) {
        if (e.lengthComputable) {
            //Loaded bytes.
            var cLoaded = e.loaded;
        }
    },
    onLoadStart: function () {
    },
    onAbort: function () {
    },
    onReadyState: function () {
        var state = xhr.readyState;
        var httpStatus = xhr.status;

        if (state === 4 && httpStatus === 200) {
            //Completed success.
            var data = xhr.responseText;
        }

    }
};
//CONTINUE YOUR CODE HERE.
xhr.open('POST', 'mypage.php', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');


if ('FormData' in window) {
    var formData = new FormData();
    formData.append("user", "aaaaa");
    formData.append("pass", "bbbbb");

    xhr.send(formData);

} else {

    xhr.send("?user=aaaaa&pass=bbbbb");
}

其他回答

只是为了让专题读者发现这个问题。我发现,只要你有一个给定的路径,接受的答案就可以工作,但如果你让它为空,它将在IE中失败。以下是我想到的:

function post(path, data, callback) {
    "use strict";
    var request = new XMLHttpRequest();

    if (path === "") {
        path = "/";
    }
    request.open('POST', path, true);
    request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
    request.onload = function (d) {
        callback(d.currentTarget.response);
    };
    request.send(serialize(data));
}

你可以这样写:

post("", {orem: ipsum, name: binny}, function (response) {
    console.log(respone);
})

不需要插件!

选择下面的代码并将其拖到书签栏中(如果你看不到它,从浏览器设置中启用),然后编辑该链接:

javascript:var my_params = prompt("Enter your parameters", "var1=aaaa&var2=bbbbb"); var Target_LINK = prompt("Enter destination", location.href); function post(path, params) { var xForm = document.createElement("form"); xForm.setAttribute("method", "post"); xForm.setAttribute("action", path); for (var key in params) { if (params.hasOwnProperty(key)) { var hiddenField = document.createElement("input"); hiddenField.setAttribute("name", key); hiddenField.setAttribute("value", params[key]); xForm.appendChild(hiddenField); } } var xhr = new XMLHttpRequest(); xhr.onload = function () { alert(xhr.responseText); }; xhr.open(xForm.method, xForm.action, true); xhr.send(new FormData(xForm)); return false; } parsed_params = {}; my_params.split("&").forEach(function (item) { var s = item.split("="), k = s[0], v = s[1]; parsed_params[k] = v; }); post(Target_LINK, parsed_params); void(0);

这是所有!现在你可以访问任何网站,并点击书签栏的按钮!


注意:

上面的方法使用XMLHttpRequest方法发送数据,因此,在触发脚本时必须处于相同的域中。这就是为什么我更喜欢用模拟表单提交发送数据,它可以将代码发送到任何域-这里是为此编写的代码:

 javascript:var my_params=prompt("Enter your parameters","var1=aaaa&var2=bbbbb"); var Target_LINK=prompt("Enter destination", location.href); function post(path, params) {   var xForm= document.createElement("form");   xForm.setAttribute("method", "post");   xForm.setAttribute("action", path); xForm.setAttribute("target", "_blank");   for(var key in params) {   if(params.hasOwnProperty(key)) {        var hiddenField = document.createElement("input");      hiddenField.setAttribute("name", key);      hiddenField.setAttribute("value", params[key]);         xForm.appendChild(hiddenField);     }   }   document.body.appendChild(xForm);  xForm.submit(); }   parsed_params={}; my_params.split("&").forEach(function(item) {var s = item.split("="), k=s[0], v=s[1]; parsed_params[k] = v;}); post(Target_LINK, parsed_params); void(0); 

尝试使用json对象而不是formdata。下面是为我工作的代码。formdata也不适合我,因此我提出了这个解决方案。

var jdata = new Object();
jdata.level = levelVal; // level is key and levelVal is value
var xhttp = new XMLHttpRequest();
xhttp.open("POST", "http://MyURL", true);
xhttp.setRequestHeader('Content-Type', 'application/json');
xhttp.send(JSON.stringify(jdata));

xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      console.log(this.responseText);
    }
}

有一些重复的作品涉及到这一点,但没有人真正阐述它。我将借用公认答案的例子来说明

http.open('POST', url, true);
http.send('lorem=ipsum&name=binny');

为了说明,我过度简化了这一点(我使用http.onload(function(){})而不是那个答案的旧方法)。如果你按原样使用,你会发现你的服务器可能会将POST正文解释为字符串,而不是实际的key=value参数(即PHP不会显示任何$_POST变量)。你必须在http.send()之前传递表单头文件来获取它。

http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');

如果您使用的是JSON而不是url编码的数据,则应该传递application/ JSON

我也遇到过类似的问题,使用相同的帖子和这个链接,我已经解决了我的问题。

 var http = new XMLHttpRequest();
 var url = "MY_URL.Com/login.aspx";
 var params = 'eid=' +userEmailId+'&amp;pwd='+userPwd

 http.open("POST", url, true);

 // Send the proper header information along with the request
 //http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
 //http.setRequestHeader("Content-Length", params.length);// all browser wont support Refused to set unsafe header "Content-Length"
 //http.setRequestHeader("Connection", "close");//Refused to set unsafe header "Connection"

 // Call a function when the state 
 http.onreadystatechange = function() {
    if(http.readyState == 4 && http.status == 200) {
        alert(http.responseText);
    }
 }
 http.send(params);

此链接已完成信息。